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प्रश्न
Present age of mother of Manish is 1 year more than 5 times the present age of Manish. Four years before, if the product of their ages was 22, then find the present age of Manish and his mother
उत्तर
Let the present age of Manish be x years.
The present age of the mother of Manish is 1 year more than 5 times the present age of Manish.
∴ The present age of the mother of Manish will be (5x + 1) years.
Four years before,
Manish’s age was (x – 4) years and
Mother’s age was (5x + 1 – 4) years i.e., (5x – 3) years
According to the given condition,
Four years before, the product of their ages was 22.
∴ (x – 4)(5x – 3) = 22
∴ 5x2 – 3x – 20x + 12 – 22 = 0
∴ 5x2 – 23x – 10 = 0
| 5 × (-10) = | -50 |
| +25 +2 | |
| (-25) × 2 = -50 | |
| -25 + 2 = -23 |
∴ 5x2 – 25x + 2x – 10 = 0
∴ 5x(x – 5) + 2(x – 5) = 0
∴ (x – 5)(5x + 2) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
x – 5 = 0 or 5x + 2 = 0
∴ x = 5 or x = `(-2)/5`
But, age cannot be negative.
∴ x = 5
∴ Manish’s present age = x = 5 years,
Mother’s present age = 5x + 1
= 5(5) + 1
= 25 + 1
= 26 years
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Solution:
We need one of the solutions of the quadratic equation as 5.
Then we can take another root as any number like a positive or negative number or zero. Here I am taking another root of the quadratic equation as 2.
Then we can form a word problem as below,
Smita is younger than her sister Mita by 3 years (5 – 2 = 3). If the product of their ages is (5 × 2 = 10). Then find their present ages.
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Therefore age of Smita = x – 3
By the given condition,
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